제출 #58976

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
58976		Benq	Brunhilda’s Birthday (BOI13_brunhilda)	C++14	5.56 / 100	907 ms	19664 KiB

brunhilda

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

template<class T, int SZ> struct Seg {
    T seg[2*SZ], MN = 0;
    
    Seg() {
        F0R(i,2*SZ) seg[i] = MOD;
    }
    
    T comb(T a, T b) { return min(a,b); } // easily change this to min or max
    
    void upd(int p, T value) {  // set value at position p
        for (seg[p += SZ] = value; p > 1; p >>= 1)
            seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
    }
    
    void build() {
        F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
    }
    
    T query(int l, int r) {  // sum on interval [l, r]
        T res1 = MN, res2 = MN; r++;
        for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
            if (l&1) res1 = comb(res1,seg[l++]);
            if (r&1) res2 = comb(seg[--r],res2);
        }
        return comb(res1,res2);
    }
};

Seg<int,1<<17> S;
int m,Q, mx, ans[1000001], p[MX];
priority_queue<pi,vpi,greater<pi>> use;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> m >> Q;
    F0R(i,m) {
        cin >> p[i];
        use.push({p[i],i});
        S.upd(i,0);
    }
    vi todo;
    FOR(i,1,1000001) {
        while (sz(use) && use.top().f == i) {
            S.upd(use.top().s,MOD);
            todo.pb(use.top().s);
            use.pop();
        }
        if (sz(use) == 0) {
            mx = i-1;
            break;
        }
        ans[i] = S.seg[1]+1;
        for (int j: todo) {
            S.upd(j,ans[i]);
            use.push({p[j]+i,j});
        }
        todo.clear();
    }
    F0R(i,Q) {
        int x; cin >> x;
        if (x > mx) cout << "oo";
        else cout << ans[x];
        cout << "\n";
    }
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 5.555555555555555 / 20
• Subtask 2: 0 / 20
• Subtask 3: 0 / 60